tag:blogger.com,1999:blog-6555947.post8434892280524982367..comments2014-01-12T10:46:48.153-07:00Comments on The Geomblog: A note on bit operations.Suresh Venkatasubramaniannoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6555947.post-58362533214501372582008-05-08T16:48:00.000-06:002008-05-08T16:48:00.000-06:00I think the argument is something like this.If we ...I think the argument is something like this.<BR/><BR/>If we weaken a class C enough, we can show that P \neq C. The point is that since we want to separate sequential from parallel, we want a "strong enough" parallel class and a "weak enough" sequential class. <BR/><BR/>Since max flow is P-complete (i.e it's strong in P), and in SP (i.e in the "weaker class"), it justifies consider SP as the correct sequential analog for PRAM-without-bitops. As he argues, P-without-bitops is meaningless as it stands. <BR/><BR/>Bottomline: using the bit extraction lower bound to say that we're done is sort of "cheating by playing games with bits". Using SP truly puts the sequential/parallel gap to the test.Sureshhttp://www.blogger.com/profile/15898357513326041822noreply@blogger.comtag:blogger.com,1999:blog-6555947.post-80891830884329912752008-05-08T13:55:00.000-06:002008-05-08T13:55:00.000-06:00I'm afraid I don't follow.Even if Mulmuley is prov...I'm afraid I don't follow.<BR/><BR/>Even if Mulmuley is proving the stronger result SP \neq PRAM-without-bitops, don't we still trivially(?) conclude from your last post that P \neq PRAM-witout-bitops?Anonymousnoreply@blogger.com